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          Luogu P3469 [POI2008]BLO-Blockade 解题报告
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        <blockquote>
<p><a href="https://www.luogu.org/problem/P3469" target="_blank" rel="noopener">[POI2008]BLO-Blockade</a></p>
<p>有 $n$ 个节点的无向图，定义封锁一个点为切断这个点的所有连边。求每个节点被封锁后图内的不连通有序点对个数。</p>
</blockquote>
<a id="more"></a>
<h2 id="解题思路："><a href="#解题思路：" class="headerlink" title="解题思路："></a>解题思路：</h2><p>Tarjan。</p>
<p>首先分类讨论一下，封锁一个点有两种情况：</p>
<ol>
<li><p>不是割点</p>
<p>这种情况好搞，从图中显然可以看出只有自己和其他 $n - 1 $ 个节点不连通，因为是有序节点，所以答案为 $2 \times (n-1)$</p>
<p><img src="https://s2.ax1x.com/2019/09/08/n3Muo8.png" alt=""></p>
</li>
<li><p>是割点</p>
<p>这种情况就有意思了。</p>
<p>我们可以发现，如果点 i 为割点，显然去掉这个点之后整个图会变成几个联通块，如下图：</p>
<p><img src="https://s2.ax1x.com/2019/09/08/n3MWFO.png" alt=""></p>
<p>这种情况我们也很好发现，把联通块的大小两两相乘可得答案。</p>
<p>记第 i 个联通块为$s_i$</p>
<p>但是把联通块大小两两相乘的复杂度为 $O(n^2)$ 不能接受，我们可以在 dfs 时把搜索树子树大小算出来，记为 $siz[i]$</p>
<p>最后的答案即为： </p>
<p>$(n - 1 - \sum_{i=1}^{t}siz[s_k])*(1+\sum_{i=1}^{t}siz[s_k])$ </p>
</li>
</ol>
<h2 id="代码："><a href="#代码：" class="headerlink" title="代码："></a>代码：</h2><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br></pre></td><td class="code"><pre><span class="line">#include &lt;cstdio&gt;</span><br><span class="line">#include &lt;cctype&gt;</span><br><span class="line">#include &lt;algorithm&gt;</span><br><span class="line">using namespace std;</span><br><span class="line">const int N &#x3D; 100010;</span><br><span class="line">const int M &#x3D; 500010&lt;&lt;1;</span><br><span class="line">inline int read() &#123;</span><br><span class="line">	int x &#x3D; 0,f &#x3D; 1;char v &#x3D; getchar();</span><br><span class="line">	while (!isdigit(v)) &#123;if (v &#x3D;&#x3D;&#39;-&#39;) f &#x3D; -1;v &#x3D; getchar();&#125;</span><br><span class="line">	while (isdigit(v)) &#123;x &#x3D; x * 10 + v - 48;v &#x3D; getchar();&#125;</span><br><span class="line">	return x * f;</span><br><span class="line">&#125;</span><br><span class="line">int nxt[M],hd[N],to[M],tot &#x3D; 1,cnt,dfn[N],low[N],siz[N],n,m;</span><br><span class="line">long long ans[N];</span><br><span class="line">bool cut[N];</span><br><span class="line"></span><br><span class="line">inline void adde(int u,int v) &#123;</span><br><span class="line">	to[++tot] &#x3D; v;nxt[tot] &#x3D; hd[u];hd[u] &#x3D; tot;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">inline void addedge(int u,int v) &#123;</span><br><span class="line">	adde(u,v);adde(v,u);</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">void tarjan(int x) &#123;</span><br><span class="line">	dfn[x] &#x3D; low[x] &#x3D; ++cnt;</span><br><span class="line">	siz[x] &#x3D; 1;</span><br><span class="line">	int flag &#x3D; 0,sum &#x3D; 0;</span><br><span class="line">	for (int i &#x3D; hd[x];i;i &#x3D; nxt[i]) &#123;</span><br><span class="line">		int v &#x3D; to[i];</span><br><span class="line">		if (!dfn[v]) &#123;</span><br><span class="line">			tarjan(v);</span><br><span class="line">			low[x] &#x3D; min(low[x],low[v]);</span><br><span class="line">			siz[x] +&#x3D; siz[v];</span><br><span class="line">			if (low[v] &gt;&#x3D; dfn[x]) &#123;</span><br><span class="line">				flag++;</span><br><span class="line">				ans[x] +&#x3D; (long long)siz[v]*(n - siz[v]);</span><br><span class="line">				sum +&#x3D; siz[v];</span><br><span class="line">				if (x !&#x3D; 1 || flag &gt; 1) &#123;</span><br><span class="line">					cut[x] &#x3D; 1;</span><br><span class="line">				&#125;</span><br><span class="line">			&#125;</span><br><span class="line">		&#125;	</span><br><span class="line">		else &#123;</span><br><span class="line">			low[x] &#x3D; min(low[x],dfn[v]);</span><br><span class="line">		&#125;</span><br><span class="line">	&#125;</span><br><span class="line">	if (cut[x]) &#123;</span><br><span class="line">		ans[x] +&#x3D; (long long)(n - sum - 1) * (sum + 1) + (n - 1);</span><br><span class="line">	&#125;</span><br><span class="line">	else &#123;</span><br><span class="line">		ans[x] &#x3D; 2*(n-1);</span><br><span class="line">	&#125;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">int main() &#123;</span><br><span class="line">	n &#x3D; read(),m &#x3D; read();</span><br><span class="line">	for (int i &#x3D; 1;i &lt;&#x3D; m;++i) &#123;</span><br><span class="line">		int x &#x3D; read(),y &#x3D; read();</span><br><span class="line">		if (x &#x3D;&#x3D; y) &#123;</span><br><span class="line">			continue;</span><br><span class="line">		&#125;</span><br><span class="line">		addedge(x,y);</span><br><span class="line">		</span><br><span class="line">	&#125;</span><br><span class="line">	tarjan(1);</span><br><span class="line">	for (int i &#x3D; 1;i &lt;&#x3D; n;++i) &#123;</span><br><span class="line">		printf(&quot;%lld\n&quot;,ans[i]);</span><br><span class="line">	&#125;</span><br><span class="line">	return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<h2 id="参考："><a href="#参考：" class="headerlink" title="参考："></a>参考：</h2><p>部分思路来自于lyd的《算法竞赛进阶指南》</p>

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